Let $K$ be the splitting field of $x^6-5$ over $\Bbb{Q}$,
(a) Let $\omega_6$ be a primitive sixth root of unity over $\Bbb{Q}$. Compute the Galois group of $K$ over $\Bbb{Q}(\omega_6)$.
(b) Compute the Galois group of $K$ over $\Bbb{Q}$.
(a) Let $\alpha = 5^{1/6}$. We know that the roots of $x^6-5$ are $\alpha$, $\alpha\omega_6$, $\alpha\omega_6^2$, $\alpha\omega_6^3$, $\alpha\omega_6^4$, and $\alpha\omega_6^5$.
Since $x^6-5$ is irreducible over $\Bbb{Q}$ by Eisenstein's criteria, $[\Bbb{Q}(\alpha, \omega_6):\Bbb{Q}] \leq 6$. We also know that
$$[\Bbb{Q}(\alpha, \omega_6):\Bbb{Q}]=[\Bbb{Q}(\alpha,\omega_6):\Bbb{Q}(\omega_6)][\Bbb{Q}(\omega_6):\Bbb{Q})].$$
Since $[\Bbb{Q}(\omega_6):\Bbb{Q}]=\phi(6)=(3-1)(2-1)=2$, $[\Bbb{Q}(\alpha,\omega_6):\Bbb{Q}(\omega_6)]=1$ or $3$. But since $\alpha \not\in \Bbb{Q}(\omega_6)$, it must be of degree 3.
So $[\Bbb{Q}(\alpha,\omega_6):\Bbb{Q}(\omega_6)] = |Gal(\Bbb{Q}(\alpha,\omega_6)/\Bbb{Q}(\omega_6))|=3$. Since there is only one group of order 3 up to isomorphism, $Gal(\Bbb{Q}(\alpha,\omega_6)/\Bbb{Q}(\omega_6)) \cong \Bbb{Z}_3$.
(b) We know that $|Gal(K/\Bbb{Q})|=6$, and there are only two groups of order 6 up to isomorphism, $\Bbb{Z}_6$ and $S_3$.
We know that there is an automorphism $\sigma: \alpha \rightarrow \alpha\omega_6$, which is cyclic of order 6, so $Gal(K/\Bbb{Q}) \cong \Bbb{Z}_6$.
Best Answer
Your first sentence beginning with the word “Since” is not correct. The right conclusion to draw from the irreducibility of $X^6-5$ over $\mathbb Q$ is that $[\mathbb Q(\alpha)\colon\mathbb Q]=6$. Now the question is whether you get a quadratic extension of $\mathbb Q(\alpha)$ by adjoining $\omega_6=\frac12(1+\sqrt{-3}\,)$, the quantity that’s missing to give you the splitting field.
I would attack the problem by following the suggested strategy and looking at $\mathbb Q(\omega_6)=k$ as your base field. I would continue and use information that maybe you don’t have, namely that the ring of integers of $k$, namely $\mathbb Z[\omega_6]$, is a principal ideal domain. And its primes are the ordinary primes congruent to $2$ modulo $3$, plus pairs of primes $\pi$, $\pi'$, one such pair for each $p\equiv1\pmod3$, and satisfying $\pi\pi'=p$. Plus a special prime $\theta$ for which $\theta^2=3$, the last of these statements being inaccurate because I suppressed mention of any unit factor. So of course, $\theta=\sqrt{-3}$ and $\theta^2$ is actually equal to $-3$. The important fact, though, is that $5$ is still prime in $\mathbb Z[\omega_6]$, so that $X^6-5$ is still Eisenstein, giving a degree of six for $K$ over $k$.The rest I think you can do.